Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{-z + 7}{-3z + 3} \div \dfrac{z^2 - 4z - 21}{z^2 - 2z - 15} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-z + 7}{-3z + 3} \times \dfrac{z^2 - 2z - 15}{z^2 - 4z - 21} $ First factor out any common factors. $n = \dfrac{-(z - 7)}{-3(z - 1)} \times \dfrac{z^2 - 2z - 15}{z^2 - 4z - 21} $ Then factor the quadratic expressions. $n = \dfrac {-(z - 7)} {-3(z - 1)} \times \dfrac {(z + 3)(z - 5)} {(z + 3)(z - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-(z - 7) \times (z + 3)(z - 5) } {-3(z - 1) \times (z + 3)(z - 7) } $ $n = \dfrac {-(z + 3)(z - 5)(z - 7)} {-3(z + 3)(z - 7)(z - 1)} $ Notice that $(z + 3)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-\cancel{(z + 3)}(z - 5)(z - 7)} {-3\cancel{(z + 3)}(z - 7)(z - 1)} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $n = \dfrac {-\cancel{(z + 3)}(z - 5)\cancel{(z - 7)}} {-3\cancel{(z + 3)}\cancel{(z - 7)}(z - 1)} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $n = \dfrac {-(z - 5)} {-3(z - 1)} $ $ n = \dfrac{z - 5}{3(z - 1)}; z \neq -3; z \neq 7 $